Continuity of a function from $R^\omega \to R^\omega$ for uniform, product and box topology.

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Bear with me, because neither my English nor my Topology is as good as it should be.

So I was studying topology and found an online list of problems, I was trying to do this one:

Let $f: R^\omega \to R^\omega$ a function such that:

$f(x_1, x_2, …)= (a_1x_1+b_1, a_2x_2+b_2, …)$ for some $a_i, b_i$ already given.

Is f continuous in the uniform, the product and the box topology?

My guess is yes, for the three.

  1. In the product topology every component is a continous function itself (a linear function).

  2. In the box topology every open of the form $(c_1, d_1)\times(c_2,d_2)\times…\times(c_n, d_n)\times…$
    will be an open after aplying f.

  3. The uniform topology, and here is where I met my limits, my guess is yes, because every open in this topology is a ball of the form $B_{\bar{\rho}}(\vec{x}, \epsilon)$ and so all the function does is move it around (translate it) so the ball will have a new center, but I'm not sure.

Could you help me out?

Best Answer

I believe your answer is correct for the product topology (indeed, products of continuous maps are always continuous in the product topology).

The answer is also correct for the box topology, but your justification does not seem correct to me. Are you implying that the preimage of any open subset of $R^\omega$ will be a product of bounded, open intervals? That's not necessarily true, in particular when some of the $a_i = 0$.

For the uniform topology, I believe your answer is false. Consider for example what happens if the $a_i = i$ and $b_i = 0$. What is the preimage of the open ball $B_{\bar{\rho}}(\vec{0}, 1/2)$ (uniform metric) under that map? Does this preimage contain any open ball (again in the uniform metric)?

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